A particle of mass $m$ follows a one-dimensional trajectory $x(t)$ in the presence of a variable force $F(x, t)$. Write down an expression for the work done by this force as the particle moves from $x\left(t_{a}\right)=a$ to $x\left(t_{b}\right)=b$. Assuming that this is the only force acting on the particle, show that the work done by the force is equal to the change in the particle's kinetic energy.

What does it mean if a force is said to be conservative?

A particle moves in a force field given by

$$F(x)=\left\{\begin{array}{cc} -F_{0} e^{-x / \lambda} & x \geqslant 0 \\ F_{0} e^{x / \lambda} & x<0 \end{array}\right.$$

where $F_{0}$ and $\lambda$ are positive constants. The particle starts at the origin $x=0$ with initial velocity $v_{0}>0$. Show that, as the particle's position increases from $x=0$ to larger $x>0$, the particle's velocity $v$ at position $x$ is given by

$$v(x)=\sqrt{v_{0}^{2}+v_{e}^{2}\left(e^{-|x| / \lambda}-1\right)}$$

where you should determine $v_{e}$. What determines whether the particle will escape to infinity or oscillate about the origin? Sketch $v(x)$ versus $x$ for each of these cases, carefully identifying any significant velocities or positions.

In the case of oscillatory motion, find the period of oscillation in terms of $v_{0}, v_{e}$, and $\lambda$. [Hint: You may use the fact that

$$\int_{w}^{1} \frac{d u}{u \sqrt{u-w}}=\frac{2 \cos ^{-1} \sqrt{w}}{\sqrt{w}}$$

for $0<w<1$.]